WBJEE · Maths · Differential Equations
If \(x \frac{d y}{d x}+y=\frac{x f(x y)}{f^{\prime}(x y)}\), then \(|f(x y)|\) is equal to
- A \(k e^{x^{2} / 2}\)
- B \(\mathrm{ke}^{\mathrm{y}^{2} / 2}\)
- C \(\mathrm{ke}^{\mathrm{x}^{2}}\)
- D \(\mathrm{ke}^{\mathrm{y}^{2}}\)
Answer & Solution
Correct Answer
(A) \(k e^{x^{2} / 2}\)
Step-by-step Solution
Detailed explanation
\(x \frac{d y}{d x}+y=\frac{x f(x y)}{f^{\prime}(x y)} \Rightarrow \frac{x d y+y d x}{d x}=\frac{x f(x y)}{f^{\prime}(x y)} \Rightarrow \frac{f^{\prime}(x y) d(x y)}{f(x y)}=x d x \Rightarrow \ln f(x y)=\frac{x^{2}}{2}+C \Rightarrow|f(x y)|=k e^{x^{2} / 2}\)
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